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Dense set

In topology, a dense set (or dense subset) of a topological space X is a subset D \subseteq X whose closure \overline{D} equals the entire space X, meaning every point in X is either an element of D or a limit point of D.^[1] This property ensures that D is "spread out" throughout X in a way that captures its topological structure. An equivalent characterization is that D intersects every nonempty open subset of X.^[2] Classic examples illustrate the concept in familiar spaces. In the real numbers \mathbb{R} equipped with the standard Euclidean topology, the set of rational numbers \mathbb{Q} is countable and dense, as between any two distinct reals there exists a rational.^[3] Likewise, the set of irrational numbers is also dense in \mathbb{R}, ensuring that irrationals fill every open interval.^[4] These examples highlight how dense sets can be both countable and uncountable while permeating the space completely. Dense sets are fundamental in analysis and topology, particularly for separable metric spaces, where the existence of a countable dense subset implies the topology has a countable basis, simplifying many constructions and proofs.^[5] They underpin approximation theorems, such as the density of polynomials in continuous functions on compact intervals (Weierstrass approximation theorem),^[6] and enable the completion of metric spaces like the rationals to the reals.^[7] In broader contexts, nowhere dense sets—subsets whose closures have empty interiors—provide contrasts, aiding the study of residual and meager sets in Baire category theorem applications.^[8]

Definitions

In topological spaces

In a topological space (X, \tau), a subset D \subseteq X is dense if its closure \overline{D} equals X.^[7] The closure \overline{D} is defined as the set of all points x \in X such that every open neighborhood U \in \tau containing x satisfies U \cap D \neq \emptyset.^[7] This formulation implies that D is "everywhere" in X, with no nonempty open set entirely avoiding D.^[9] An equivalent characterization is that D intersects every nonempty open set in \tau.^[7] Specifically, for any U \in \tau with U \neq \emptyset, it holds that U \cap D \neq \emptyset.^[7] This intersection property underscores the notion of density without relying on the closure operator, emphasizing the pervasive presence of D across the space.^[9] The space X itself is always dense in X, as \overline{X} = X.^[7] Moreover, if D is dense and D \subseteq E \subseteq X, then E is also dense, since \overline{E} \supseteq \overline{D} = X implies \overline{E} = X.^[7] These properties highlight the hereditary nature of density under inclusion for subsets containing a dense set.

In metric spaces

In a metric space (X, d), a subset D \subseteq X is dense if its closure \overline{D} = X, meaning every point in X is either in D or a limit point of D.^[10] Equivalently, D is dense if for every x \in X and every \epsilon > 0, the open ball B(x, \epsilon) = \{ y \in X \mid d(x, y) < \epsilon \} intersects D.^[11] This metric-specific characterization leverages the uniformity of open balls to ensure D comes arbitrarily close to every point in the space.^[12] Another equivalent condition is that every point x \in X is the limit of a sequence \{x_n\} \subseteq D with x_n \to x.^[10] In complete metric spaces, dense subsets play a key role in extensions and completions: any metric space is isometrically embeddable as a dense subset of a complete metric space, known as its completion.^[13] For instance, the rational numbers \mathbb{Q} form a dense subset of the real numbers \mathbb{R} under the standard metric, as sequences of rationals converge to any real.^[11] A metric space X is separable if it admits a countable dense subset, which implies the existence of a countable basis for the topology generated by the metric.^[14] Compact metric spaces are always separable, as their total boundedness allows construction of a countable dense set via finite covers of small balls.^[11] In complete metric spaces without isolated points, the Baire category theorem asserts that the countable intersection of open dense sets is itself dense (and nonempty).^[15] This theorem underscores the "largeness" of dense sets in complete spaces: for example, no countable dense subset can be a G_\delta set in such spaces.^[11] Moreover, in separable metric spaces, the Borel \sigma-algebra is generated by countable unions and intersections of open balls with rational radii centered at points of the dense subset.^[14]

Examples

In the real line

In the real line \mathbb{R} equipped with the standard topology, a subset D \subseteq \mathbb{R} is dense if its closure is all of \mathbb{R}, or equivalently, if every nonempty open interval (a, b) with a < b contains at least one point from D. This property ensures that D "fills" \mathbb{R} in a topological sense, with points arbitrarily close to any real number.^[16] The rational numbers \mathbb{[Q](/page/Q)} form a countable dense subset of \mathbb{R}. For any open interval (a, b), there exists a rational p/q (with p \in \mathbb{Z}, q \in \mathbb{N}) such that a < p/q < b, as the rationals are order-dense in the reals. This density implies that \mathbb{R} is a separable metric space.^[16]^[4] The irrational numbers \mathbb{R} \setminus \mathbb{Q} are also dense in \mathbb{R}. To see this, consider any open interval (a, b); it contains a rational r, and adding a scaled irrational like \sqrt{2}/n (for sufficiently large n \in \mathbb{N}) yields an irrational r + \sqrt{2}/n \in (a, b), since the sum of a rational and a nonzero irrational is irrational. Thus, every open interval intersects the irrationals.^[4] Other examples include the algebraic numbers, which are dense in \mathbb{R} due to their inclusion of all rationals and additional dense layers from roots of polynomials, though they remain countable. In contrast, closed intervals like [0, 1] are not dense in \mathbb{R}, as they miss points outside (0, 1). These examples highlight how density in \mathbb{R} often relies on the interplay between countable and uncountable structures within the ordered field.^[4]

In other spaces

In Euclidean spaces of dimension greater than one, such as \mathbb{R}^n with the standard Euclidean topology, the set \mathbb{Q}^n consisting of all points with rational coordinates is dense. This density arises because \mathbb{Q} is dense in \mathbb{R}, and the product of dense subsets is dense in the product topology, which coincides with the Euclidean topology on \mathbb{R}^n.^[17] The complex plane \mathbb{C}, topologically equivalent to \mathbb{R}^2, provides a similar example: the set of Gaussian rationals \{a + bi \mid a, b \in \mathbb{Q}\} is dense in \mathbb{C} under the standard metric topology. Every open disk in \mathbb{C} contains Gaussian rationals, ensuring that the closure of this set is the entire plane.^[17] In spaces of continuous functions, such as C([a, b], \mathbb{R}) equipped with the supremum norm topology, the set of real polynomials is dense. The Stone–Weierstrass theorem guarantees this, as polynomials form a unital subalgebra of C([a, b], \mathbb{R}) that separates points and vanishes nowhere. For instance, on the compact interval [0, 1], any continuous function can be uniformly approximated by polynomials to arbitrary precision.^[18]

Properties

Closure properties

In a topological space X, a subset D \subseteq X is dense if its closure \overline{D} = X. Consequently, the closure of any dense set coincides with the entire space X.^[7]^[19] The closure operator satisfies monotonicity: if D \subseteq A \subseteq X, then \overline{D} \subseteq \overline{A}. Thus, if D is dense in X, any superset A of D is also dense in X, as \overline{A} \supseteq \overline{D} = X.^[7]^[19] A dense set need not be closed. However, if a dense set D is closed, then D = \overline{D} = X, so it equals the whole space.^[7] The space X itself is always dense in X.^[7]

Intersection and unions

In a topological space X, the union of any family of dense subsets is dense. To see this, consider a nonempty open set U \subseteq X. Since each set in the family is dense, U intersects every member of the family, so U intersects their union.$$] This holds for both finite and infinite unions, as the property relies only on the individual density of each set. In contrast, the intersection of dense subsets need not be dense. For instance, in the real line \mathbb{R} with the standard topology, both the rational numbers \mathbb{Q} and the irrational numbers \mathbb{R} \setminus \mathbb{Q} are dense, but \mathbb{Q} \cap (\mathbb{R} \setminus \mathbb{Q}) = \emptyset, and the empty set is not dense.[$$ Arbitrary intersections of dense sets can fail to be dense even in simpler cases, such as when the sets are not open. However, the situation improves when the dense sets are open. The intersection of finitely many open dense subsets of X is open and dense. For two such sets U and U', if V \subseteq X is nonempty and open, then V \cap U is nonempty and open (hence intersects U'), so V \cap (U \cap U') \neq \emptyset; this extends inductively to finite collections.

\] In Baire spaces, such as complete metric spaces, even countable intersections of open dense sets remain dense.\[

Nowhere dense sets

A subset A of a topological space X is nowhere dense if the interior of its closure is empty, that is, \operatorname{int}(\overline{A}) = \emptyset.^[20] This condition is equivalent to the closure \overline{A} containing no nonempty open subset of X.^[20] An alternative characterization is that for every nonempty open set U \subseteq X, there exists a nonempty open set V \subseteq U such that V \cap A = \emptyset.^[21] Nowhere dense sets contrast sharply with dense sets: while a dense set A satisfies \overline{A} = X (so its closure fills the entire space), a nowhere dense set's closure avoids having any substantial "thickness" by excluding open neighborhoods entirely.^[20] The complement of a nowhere dense set is dense in X, meaning it intersects every nonempty open set.^[20] This notion captures sets that are "sparse" everywhere, with no region where they dominate topologically. In metric spaces, closed nowhere dense sets provide key examples. For instance, any finite subset of \mathbb{R} is nowhere dense, as its closure is itself and has empty interior.^[21] The middle-thirds Cantor set C \subseteq [0,1] is a canonical example of a closed, perfect (dense-in-itself with no isolated points), nowhere dense set in \mathbb{R}, as its closure is itself and contains no intervals (hence empty interior).^[21] In contrast, the rationals \mathbb{Q} \subseteq \mathbb{R} are not nowhere dense, since \overline{\mathbb{Q}} = \mathbb{R} and \operatorname{int}(\mathbb{R}) = \mathbb{R} \neq \emptyset.^[20] Nowhere dense sets play a central role in Baire category theory. A set is meager (or of first category) if it is a countable union of nowhere dense sets; otherwise, it is of second category.^[20] The Baire category theorem asserts that complete metric spaces (and more generally, Baire spaces) cannot be written as a countable union of nowhere dense sets, implying they are of second category.^[21] For example, \mathbb{R} is not a countable union of nowhere dense sets, which has implications like the uncountability of closed intervals and Cantor sets.^[21] Properties of nowhere dense sets include closure invariance: if A is nowhere dense, then \overline{A} is also nowhere dense.^[20] Finite unions of nowhere dense sets are nowhere dense, but countable unions need not be (e.g., \mathbb{Q} as a countable union of singletons).^[21] In normal spaces, for any nowhere dense M and nonempty open U, there exists open V \subseteq U with V \cap M = \emptyset and \overline{V} \cap M = \emptyset.^[21] Boundaries of open sets are nowhere dense, underscoring their role in describing "thin" topological features.^[21]

Residual sets

In topology, a residual set (also called a comeager set) in a topological space is defined as the complement of a meager set, where a meager set is a countable union of nowhere dense subsets.^[22] Equivalently, a set is residual if it contains a countable intersection of open dense subsets.^[15] This notion arises prominently in the context of the Baire category theorem, which asserts that in a complete metric space (a Baire space), the countable intersection of open dense sets is itself dense.^[23] Consequently, every residual set in such a space is dense, providing a stronger "largeness" property in the sense of category compared to mere density.^[24] Residual sets are of second category, meaning they are not meager, and their complements are "small" in the topological category sense.^[25] Unlike dense sets, which have closures equal to the entire space but may be meager (e.g., the rational numbers \mathbb{Q} in \mathbb{R}), residual sets capture sets that are generically large under the Baire category framework.^[26] For instance, in the real line \mathbb{R} with the standard topology, the set of irrational numbers \mathbb{R} \setminus \mathbb{Q} is residual because its complement \mathbb{Q} is meager (a countable union of singletons, each nowhere dense).^[15] This example illustrates how residual sets often coincide with dense G_\delta sets in complete metric spaces, as the countable intersection of open dense sets yields a dense G_\delta set that is residual.^[27] The concept is particularly useful in descriptive set theory and functional analysis, where residual sets help characterize generic properties of functions or points in infinite-dimensional spaces, such as the set of continuous functions in certain topologies being residual.^[28] In non-Baire spaces, residual sets need not be dense, highlighting the theorem's role in ensuring their ubiquity in "nice" topological settings.^[29]

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